Consider a 0.487 L solution of the amino aspartic acid (0.685 M), which ahs a carboxylic...

Consider a 0.487 L solution of the amino aspartic acid (0.685 M), which ahs a carboxylic acid group (pKa = 2.10), an amine group (pKa = 9.82), and a carboxylic acid side chain (pKa = 3.86). How many liters of 2.59 M NaOH would you need to add to reach the isoelectric point of the amino acid?

Expert Solution

ANS) given side chain Pka =3.86 ,acidic group Pka = 2.10

isoelectric point ( PH) = 3.86 + 2.10 / 2 = 2.98

NaOH + ASPARTIC ACID ------------ SALT + H2O

mole added 2.59* V 0.487 * 0.685=0.3335 0 0

moles after rxn 0 0.3335 - 2.59V 2.59V 2.59V

[ ACID ] = 0.3335 -2.59V / 0.487+ V , [ SALT ] = 2.59V / 0.487+ V

USING Hendersons eq P^H = P^Ka^{+ log [ salt ]} / [acid ]

2.98 = 2.10 + log [ 2.59V / 0.487+ V] / [ 0.3335 - 2.59V / 0.487 +V

0.88 = log 2.59v+ 0.4749 + log 2.59v

v =0.2364 L

thank u

queen_honey_blossom answered 2 years ago

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